In Apostol Calculus 1, the complex logarithm is defined as follows:
$$\text{Log}(z) = \log(|z|) + i \; \text{arg}(z)$$
We assume $\text{arg}(z) \in (-\pi, \pi]$. However, in the next exercise, we need to show that:
$$\text{Log}(z_1z_2) = \text{Log}(z_1) + \text{Log}(z_2) + 2n\pi i, n \in Z$$
I do not see how he gets to this outcome using the definition:
$$\text{Log}(z_1z_2) = \log(|z_1||z_2|) + i \; \text{arg}(z_1z_2) = \log(|z_1|) +\log(|z_1|) + i \; \big(\text{arg}(z_1) + \text{arg}(z_2)\big)$$
or $\text{Log}(z_1z_2) = \text{Log}(z_1) + \text{Log}(z_2)$. Where is the summand $2n\pi i$ supposed to appear from?
This detail necessary to show the restrictions on the following fact:
$$(z_1z_2)^w = z_1^wz_2^w$$
where this summand plays an important role to show the domain of the complex numbers $z_1$ and $z_2$.
The point is $\arg(z_1z_2)\neq\arg(z_1)+\arg(z_2)$, since the argument function is valued in a finite interval: $[0,2\pi)$, usually.
Thus if $z_1=z_2=-i$, $z_1z_2=-1$ and
$$\arg(z_1z_2)=\pi\neq 3\pi=\arg(z_1)+\arg(z_2)$$
What is true is that the argument of the product of two complex numbers is either the sum of the arguments or the sum plus $2\pi$. Hence, generalizing to any number of numbers, the formula holds up to adding an entire number of full angles, i.e. $2n\pi$.